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Elementary Algebra

Katherine Yoshiwara

Section 7.3 More About Factoring

Subsection Quotients of Powers

We can reduce a fraction by dividing numerator and denominator by any common factors. For example,
\begin{equation*} \dfrac{10}{15} = \dfrac{2 \cdot 5}{3 \cdot 5} = \dfrac{2}{3} \end{equation*}
We first factored the numerator and denominator of the fraction and then canceled the 5’s by dividing. We can apply the same technique to quotients of powers.

Example 7.20.

Simplify.
  1. \(\displaystyle \dfrac{a^5}{a^3}\)
  2. \(\displaystyle \dfrac{a^4}{a^8}\)
Solution.
  1. We first write the numerator and denominator in factored form. Then we divide any common factors from the numerator and denominator.
    \begin{equation*} \dfrac{a^5}{a^3} = \dfrac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a\cdot a}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}} = \dfrac{a^2}{1} = a^2 \end{equation*}
    You may observe that the exponent of the quotient can be obtained by subtracting the exponent of the denominator from the exponent of the numerator. In other words,
    \begin{equation*} \dfrac{a^5}{a^3}= a^{5-3} = a^2 \end{equation*}
  2. In this quotient, the larger power occurs in the denominator.
    \begin{equation*} \dfrac{a^4}{a^8} = \dfrac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot \cancel{a}}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a\cdot a\cdot a \cdot a} = \dfrac{1}{a^4} \end{equation*}
    We see that we can subtract the exponent of the numerator from the exponent of the denominator. That is,
    \begin{equation*} \dfrac{a^4}{a^8}=\dfrac{1} {a^{8-4}} = \dfrac{1}{a^4} \end{equation*}
These examples suggest the following property.

Second Law of Exponents.

To divide two powers with the same base, we subtract the smaller exponent from the larger one, and keep the same base.
  1. If the larger exponent occurs in the numerator, put the power in the numerator.
  2. If the larger exponent occurs in the denominator, put the power in the denominator.
In symbols,
  1. \(\displaystyle \blert{\dfrac{a^m}{a^n} = a^{m-n}~~~~~~\text{if}~~~~n \lt m}\)
  2. \(\displaystyle \blert{\dfrac{a^m}{a^n} = \dfrac{1}{a^{n-m}}~~~~~~\text{if}~~~~n \gt m}\)

Reading Questions Reading Questions

1.
In a quotient of powers, how do we know whether the new power appears in the numerator or the denominator?
Answer.
By which has the larger exponent

Subsection Quotients of Monomials

To divide one monomial by another, we apply the second law of exponents to the powers of each variable.

Example 7.21.

Divide \(~~\dfrac{3x^2y^4}{6x^3y}\)
Solution.
We consider the numerical coefficients and the powers of each base separately. We use the second law of exponents to simplify each quotient of powers.
\begin{align*} \dfrac{3x^2y^4}{6x^3y} \amp = \dfrac{3}{6} \cdot \dfrac{x^2}{x^3} \cdot \dfrac{y^4}{y} \amp \amp \blert{\text{Subtract exponents on each base.}}\\ \amp = \dfrac{1}{2} \cdot \dfrac{1}{x^{3-2}} \cdot y^{4-1}\\ \amp = \dfrac{1}{2} \cdot \dfrac{1}{x} \cdot \dfrac{y^3}{1}=\dfrac{y^3}{2x} \amp \amp \blert{\text{Multiply factors.}} \end{align*}

Subsection Greatest Common Factors

Now we consider several techniques for factoring polynomials. The first of these is factoring out the greatest common factor (GCF).

Greatest Common Factor.

The greatest common factor (GCF) is the largest factor that divides evenly into each term of the polynomial: the largest numerical factor and the highest power of each variable.

Example 7.22.

Find the greatest common factor for \(~~4a^3b^2+6ab^3-18a^2b^4\)
Solution.
The largest integer that divides evenly into all three coefficients is 2. The highest power of \(a\) is \(a^1\text{,}\) and the highest power of \(b\) is \(b^2\text{.}\) Thus, the GCF is \(2ab^2\text{.}\) Note that the exponent on each variable of \(2ab^2\) is the smallest exponent that appears on that variable among the terms of the polynomial.

Reading Questions Reading Questions

2.
The exponent on each variable of the GCF is the exponent that appears on that variable among the terms of the polynomial.
Answer.
smallest
Once we have found the greatest common factor for the polynomial, we can write each term as a product of the GCF and another factor. For example, the GCF of \(8x^2-6x\) is \(2x\text{,}\) and we can write
\begin{equation*} 8x^2-6x = \blert{2x} \cdot 4x - \blert{2x} \cdot 3 \end{equation*}
We can then use the distributive law to write the expression on the right side as a product.
\begin{equation*} \blert{2x} \cdot 4x - \blert{2x} \cdot 3 = \blert{2x} (4x-3) \end{equation*}
We say that we have factored out the greatest common factor from the polynomial.
For more complicated polynomials, we can divide the GCF into each term to find the remaining factors.

Example 7.23.

Factor \(~~4a^3b^2+6ab^3-18a^2b^4\)
Solution.
The GCF for this polynomial is \(2ab^2\text{,}\) as we saw in Example 7.22. We factor out the GCF from each term and write the polynomial as a product,
\begin{equation*} 2ab^2(\hphantom{0000000000}) \end{equation*}
To find the factor inside parentheses, we divide each term of the polynomial by the GCF.
\begin{equation*} \dfrac{4a^3b^2}{2ab^2}=\blert{2a^2},~~~~\dfrac{6ab^3}{2ab^2}=\blert{3b},~~~\dfrac{-18a^2b^4}{2ab^2}=\blert{-9ab^2} \end{equation*}
We apply the distributive law to factor \(2ab^2\) from each term.
\begin{align*} 4a^3b^2+6ab^3-18a^2b^4 \amp =2ab^2\cdot \blert{2a^2}+2ab^2 \cdot \blert{3b-}2ab^2 \cdot \blert{9ab^2}\\ \amp = 2ab^2(2a^2+3b-9ab^2) \end{align*}

Look Closer.

Sometimes the greatest common factor for a polynomial is not a monomial, but may instead have two or more terms.

Example 7.24.

Factor \(x^2(2x+1)-3(2x+1)\)
Solution.
The given expression has two terms, and \((2x+1)\) is a factor of each. We factor out the entire binomial \((2x+1)\text{.}\)
\begin{equation*} x^2\blert{(2x+1)}-3\blert{(2x+1)} = \blert{(2x+1)} (x^2-3) \end{equation*}

Subsection Factoring Quadratic Trinomials by Guess-and-Check

So far, we can factor quadratic trinomials of the form \(x^2+bx+c=0\text{,}\) where \(a=1\text{.}\) Recall that the factored form of \(x^2+bx+c=0\) looks like
\begin{equation*} (x+p)(x+q)~~~~~~ \text{where}~~~pq=c ~~~ \text{and} ~~~p+q=b \end{equation*}
You may want to review the FOIL method discussed in Lesson 6.3.
What if the coefficient of \(a\) is not 1? Sometimes, if the coefficients are small, we can factor a quadratic expression by the guess-and-check method.

Example 7.25.

Factor \(~2x^2-7x+3~\) into a product of two binomials.
Solution.
We begin by factoring the quadratic term, \(2x^2\text{,}\) which can only be factored as \(x\) times \(2x\text{,}\) so we can fill in the First terms in each binomial.
\begin{equation*} 2x^2-7x+3 = (2x\fillinmath{XXXXXX})(x\fillinmath{XXXXXX}) \end{equation*}
Next, we factor the constant term, \(3\text{,}\) which can only be factored as 3 times 1. Because the linear term is negative, we make both factors negative: \(3=-3(-1)\text{.}\) Finally, we have to decide which number appears as the Last term in each binomial. We check \(O+I\) for each possibility.
\begin{align*} (2x-3)(x-1)~~~~O+I \amp = -2x-3x=-5x\\ (2x-1)(x-3)~~~~\blert{O+I} \amp \blert{= -6x-x=-7x} \end{align*}
The second choice gives the correct middle term, so the factorization is
\begin{equation*} 2x^2-7x+3 = (2x-1)(x-3) \end{equation*}

Reading Questions Reading Questions

3.
When we factor a quadratic trinomial by the guess-and-check method, how do we check for the correct middle term?
Answer.
We compute \(O+I\text{.}\)
Remember that you can always check your factorization by multiplying the factors together.

Subsection Quadratic Trinomials in Two Variables

So far, we have factored quadratic trinomials in one variable, that is, polynomials of the form
\begin{equation*} ax^2+bx+c \end{equation*}
The method we learned can also be used to factor trinomials in two variables of the form
\begin{equation*} ax^2+bxy+cy^2 \end{equation*}
In this expression, the first and last terms are quadratic terms, while the middle term is a cross-term consisting of the product of the two variables.

Example 7.26.

Factor \(~x^2+5xy+6y^2~\)
Solution.
As usual, we begin by factoring the first term into \(x\) times \(x\text{.}\)
\begin{equation*} x^2+5xy+6y^2 = (x + \fillinmath{XXXXXX})(x + \fillinmath{XXXXXX}) \end{equation*}
Next we look for factors of the last term, \(6y^2\text{.}\) In order to obtain the \(xy\)-term in the middle, we need a \(y\) in each factor. Thus the possibilities are
\begin{equation*} y~~\text{and}~~6y~~~~~~~~\text{or}~~~~~~~~2y~~\text{and}~~3y \end{equation*}
We’ll check the sum \(O+I\) for each possibility.
\begin{align*} (x+y)(x+6y)~~~~O+I \amp = 6xy+xy=7xy\\ (x+2y)(x+3y)~~~~\blert{O+I} \amp \blert{= 3xy+2xy=5xy} \end{align*}
The second possibility gives the correct middle term, so the factorization is
\begin{equation*} x^2+5xy+6y^2 = (x+2y)(x+3y) \end{equation*}

Subsection Combining Factoring Techniques

We should always begin factoring by checking to see if there is a common factor that can be factored out.

Example 7.27.

Factor completely \(~2b^3+8b^2+6b~\)
Solution.
We begin by factoring out the greatest common factor, \(2b\text{.}\)
\begin{equation*} 2b^3+8b^2+6b = 2b(b^2+4b+3) \end{equation*}
The remaining factor, \(b^2+4b+3\text{,}\) is a quadratic trinomial that can be factored. We look for two numbers \(p\) and \(q\) so that \(pq=3\) and \(p+q=4\text{.}\) You can check that \(p=3\) and \(q=1\) will work.
Thus, \(~~b^2+4b+3 = (b+3)(b+1)\text{,}\) and
\begin{equation*} 2b^3+8b^2+6b = 2b(b+3)(b+1) \end{equation*}

Reading Questions Reading Questions

4.
What should always be the first step in factoring a polynomial?
Answer.
Factor out any common factors.

Subsection \(\blert{\text{Optional Extension: The Box Method}}\)

If the coefficients in a quadratic trinomial \(ax^2+bx+c\) are not prime numbers, the guess-and-check method may be time-consuming. In that case, we can use another technique that depends upon the following property of binomial products.

Example 7.28.

  1. Compute the product \((3t+2)(t+3)\) using the area of a rectangle.
  2. Verify that the products of the diagonal entries are equal.
Solution.
  1. We construct a rectangle with sides \(3t+2\) and \(t+3\text{,}\) as shown below. We see that the product of the two binomials is
    \(3t^2+9t+2t+6=3t^2+11t+6\)
    \(~~~~~t~~~~~\) \(~~~~3~~~~~\)
    \(3t\) \(3t^2\) \(9t\)
    \(2\) \(2t\) \(6\)
  2. Now let’s compute the product of the expressions along each diagonal of the rectangle:
    \begin{equation*} 3t^2 \cdot 6 = 18t^2~~~~~~\text{and}~~~~~~9t \cdot 2t = 18t^2 \end{equation*}
    The two products are equal. This is not surprising when you think about it, because each diagonal product is the product of all four terms of the binomials, namely \(3t,~2,~t,\) and \(3\text{,}\) just multiplied in a different order. You can see where the diagonal entries came from in our example:
    \begin{align*} 18t^2 \amp = 3t^2 \cdot 6 =3t \cdot t \cdot 2 \cdot 3\\ 18t^2 \amp = 9t \cdot 2t =3t \cdot 3 \cdot 2 \cdot t \end{align*}

Product of Binomials.

When we represent the product of two binomials by the area of a rectangle, the products of the entries on the two diagonals are equal.

Reading Questions Reading Questions

5.
In the previous Example, why are the products on the two diagonals equal?
Answer.
Each is the product of all four terms of the binomials.

Look Ahead.

We can use rectangles to help us factor quadratic trinomials. Recall that factoring is the opposite or reverse of multiplying, so we must first understand how multiplication works.

Look Closer.

Look carefully at the rectangle for the product
\begin{equation*} (3x+4)(x+2)=3x^2+10x+8 \end{equation*}
Shown at right.
\(~~~~~x~~~~~\) \(~~~~2~~~~~\)
\(3x\) \(3x^2\) \(6x\)
\(4\) \(4x\) \(8\)
  • The quadratic term of the product, \(3x^2\text{,}\) appears in the upper left sub-rectangle.
  • The constant term of the product, 8, appears in the lower right sub-rectangle.
  • The linear term, \(10x\text{,}\) is the sum of the other two sub-rectangles.

Subsection Factoring Quadratic Trinomials by the Box Method

Now we’ll factor the trinomial
\begin{equation*} 3x^2+10x+8 \end{equation*}
We’ll try to reverse the steps for multiplication. Instead of starting with the factors on the outside of the rectangle, we begin by filling in the areas of the sub-rectangles.
\(\blert{\text{Step 1}~~}\) The quadratic term, goes in the upper left, and the constant term, 8, goes in the lower right, as shown in the figure.
\(\hphantom{0000}\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(3x^2\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(\hphantom{0000}\) \(8\)
What about the other two entries? We know that their sum must be \(10x\text{,}\) but we don’t know what expressions go in each! This is where we use our observation that the products on the two diagonals are equal.
\(\blert{\text{Step 2}~~}\) We compute the product of the entries on the first diagonal:
\begin{equation*} D=3x^2 \cdot 8 = 24x^2 \end{equation*}
The product of the entries on the other diagonal must also be \(24x^2\text{.}\) We now know two things about those entries:
\begin{align*} \amp \text {1. Their product is}~~24x^2~~\text{and}\\ \amp \text {2. Their sum is}~~10x \end{align*}
\(\blert{\text{Step 3}~~}\) To find the two unknown entries, we list all the ways to factor \(D=24x^2\text{,}\) then choose the factors whose sum is \(10x\text{.}\)
Factors of \(D=24x^2\) \(\hphantom{0000}\)Sum of Factors
\(x\) \(24x\) \(\hphantom{0000}x+24x=25x\)
\(2x\) \(12x\) \(\hphantom{0000}2x+12x=24x\)
\(3x\) \(8x\) \(\hphantom{0000}3x+8x=11x\)
\(\blert{4x}\) \(\blert{6x}\) \(\hphantom{0000}\blert{4x+6x=10x}\)
\(\blert{\text{Step 4}~~}\) We see that the last pair of factors, \(4x\) and \(6x\text{,}\) has a sum of \(10x\text{.}\) We enter these factors in the remaining sub-rectangles. (It doesn’t matter which one goes in which spot.) We now have all the sub-rectangles filled in, as shown at right.
\(\) \(\)
\(\) \(~~~3x^2~~~\) \(~~~4x~~~\)
\(\) \(~~~6x~~~\) \(~~~8~~~\)
Finally, we work backwards to discover what length and width produce the areas of the four subrectangles.
\(\blert{\text{Step 5}~~}\) We factor each row of the rectangle, and write the factors on the outside. Start with the top row, factoring out \(x\) and writing the result, \(3x+4\text{,}\) at the top, as shown at right. We get the same result when we factor \(2\) from the bottom row.
The final rectangle is shown at right, and the factors of \(3x^2+10x+8\) appear as the length and width of the rectangle. Our factorization is thus
\begin{equation*} 3x^2+10x+8=(x+2)(3x+4) \end{equation*}
\(3x\) \(4\)
\(x\) \(\hphantom{00}3x^2\hphantom{00}\) \(\hphantom{00}4x\hphantom{00}\)
\(2\) \(\hphantom{00}6x\hphantom{00}\) \(\hphantom{00}8\hphantom{00}\)

Reading Questions Reading Questions

6.
Which terms of the quadratic trinomial go into the upper left and lower right sub-rectangles of the box?
Answer.
The quadratic and constant terms
7.
Why do we list the possible factors of \(D\text{?}\)
Answer.
To see which sum of factors equals the linear term

Example 7.29.

Factor \(~~2x^2-11x+15\)
Solution.
\(\blert{\text{Step 1}~~}\) Enter \(2x^2\) and \(15\) on the diagonal of the rectangle, as shown in the figure.
\(\hphantom{0000}\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(2x^2\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(\hphantom{0000}\) \(15\)
\(\blert{\text{Step 2}~~}\) Compute the diagonal product:
\begin{equation*} D=2x^2 \cdot 15 = 30x^2 \end{equation*}
\(\blert{\text{Step 3}~~}\) List all possible factors of \(D\text{,}\) and compute the sum of each pair of factors. (Note that both factors must be negative.)
Factors of \(D=30x^2\) \(\hphantom{0000}\)Sum of Factors
\(-x\) \(-30x\) \(\hphantom{0000}-x-30x=-31x\)
\(-2x\) \(-15x\) \(\hphantom{0000}-2x-15x=-17x\)
\(-3x\) \(-10x\) \(\hphantom{0000}-3x-10x=-13x\)
\(\blert{-5x}\) \(\blert{-6x}\) \(\hphantom{0000}\blert{-5x-6x=-11x}\)
The correct factors are \(-5x\) and \(-6x\text{.}\)
\(\blert{\text{Step 4}~~}\) Enter the factors \(-5x\) and \(-6x\) into the rectangle.
\(\hphantom{0000}\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(~~~2x^2~~~\) \(~~~-6x~~~\)
\(\hphantom{0000}\) \(~~~-5x~~~\) \(~~~15~~~\)
\(\blert{\text{Step 5}~~}\) Factor \(2x\) from the top row of the rectangle, and write the result, \(x-3\text{,}\) at the top, as shown below.
\(x\) \(-3\)
\(2x\) \(~~~2x^2~~~\) \(~~~-6x~~~\)
\(\hphantom{00}\) \(~~~-5x~~~\) \(~~~15~~~\)
\(x\) \(-3\)
\(2x\) \(~~~2x^2~~~\) \(~~~-6x~~~\)
\(-5\) \(~~~-5x~~~\) \(~~~15~~~\)
Finally, factor \(5\) from the bottom row, and write it on the left. The correct factorization is
\begin{equation*} 2x^2-11x+15=(2x-5)(x-3) \end{equation*}

Reading Questions Reading Questions

8.
What do we do after we have filled in all the sub-rectangles of the box?
Answer.
Factor the rows
9.
Where do the factors of the quadratic trinomial appear?
Answer.
As the length and width of the rectangle
Here is a summary of our factoring method.

To Factor \(~ax^2+bx+c~\) Using the Box Method.

  1. Write the quadratic term \(ax^2\) in the upper left sub-rectangle, and the constant term \(c\) in the lower right.
  2. Multiply these two terms to find the diagonal product, \(D\text{.}\)
  3. List all possible factors \(px\) and \(qx\) of \(D\text{,}\) and choose the pair whose sum is the linear term, \(bx\text{,}\) of the quadratic trinomial.
  4. Write the factors \(px\) and \(qx\) in the remaining sub-rectangles.
  5. Factor each row of the rectangle, writing the factors on the outside. These are the factors of the quadratic trinomial.

Subsection Skills Warm-Up

Exercises Exercises

Exercise Group.
Mental exercise: Find the other factor of each quadratic trinomial without using pencil, paper, or calculator.
1.
\(b^2+8b-240 = (b-12)(\fillinmath{XXXXXX})\)
2.
\(n^2-97n-300 = (n-100)(\fillinmath{XXXXXX})\)
3.
\(3u^2-17u-6 = (u-6)(\fillinmath{XXXXXX})\)
4.
\(2t^2-21t+54 = (t-6)(\fillinmath{XXXXXX})\)
Exercise Group.
\(\blert{\text{(Optional)}~~}\) Use the given areas to find the length and width of each rectangle.
5.
\(~~6x^2~~\) \(~~9x~~\)
\(~~10x~~\) \(~~15~~\)
6.
\(~~8t^2~~\) \(~~-14t~~\)
\(~~-12t~~\) \(~~21~~\)
7.
\(~~12m^2~~\) \(~~-10m~~\)
\(~~30m~~\) \(~~-25~~\)
8.
\(~~9a^2~~\) \(~~21a~~\)
\(~~-21a~~\) \(~~-49~~\)

Subsubsection Answers to Skills Warm-Up

  1. \(\displaystyle b+20\)
  2. \(\displaystyle n+3\)
  3. \(\displaystyle 3u+1\)
  4. \(\displaystyle 2t-9\)
  5. \(\displaystyle 3x+5,~2x+3\)
  6. \(\displaystyle 2t-3,~4t-7\)
  7. \(\displaystyle 2m+5,~6m-5\)
  8. \(\displaystyle 3a-7,~3a+7\)

Subsection Lesson

Subsubsection Activity 1: Quotients and the GCF

Exercises Exercises
1.
Use the first second of exponents to find each quotient.
  1. \(\displaystyle \dfrac{x^6}{x^2}\)
  2. \(\displaystyle \dfrac{b^2}{b^3}\)
2.
Divide \(~~\dfrac{8x^2y}{12x^5y^3}\)
3.
Find the greatest common factor for \(~~15x^2y^2-12xy+6xy^3\)
4.
Factor \(~~15x^2y^2-12xy+6xy^3\)
5.
Factor \(~~9(x^2+5)-x(x^2+5)\)

Subsubsection Activity 2: Quadratic Trinomials

Exercises Exercises
1.
Factor \(~3x^2+2x-5~\) by the guess-and-check method.
2.
  1. Compute the product \(~(2x-5)(3x-4)~\) using the area of a rectangle.
  2. Verify that the products of the diagonal entries are equal.
3.
Solve \(~2x^2=7x+15~\)
\(\blert{\text{Write the equation in standard form.}}\)
\(\blert{\text{Factor the left side.}}\)
\(\blert{\text{Set each factor equal to zero.}}\)
\(\blert{\text{Solve each equation.}}\)

Subsubsection Activity 3: Factoring Completely

Exercises Exercises
1.
Factor completely \(~4a^6-10a^5+6a^4\)
2.
Factor completely.
  1. \(\displaystyle 2a^3b-24a^2b^2-90ab^3\)
  2. \(\displaystyle 3x^2-8xy+4y^2\)

Subsubsection
\(\blert{\text{(Optional Extension)}~~}\) Activity 4: The Box Method

Follow the steps to factor \(~4x^2+4x-3\text{.}\)
\(\blert{\text{Step 1}~~}\) Enter the correct terms on the diagonal.
\(\blert{\text{Step 2}~~}\) Compute the diagonal product:
\begin{equation*} D= \end{equation*}
boxes
\(\blert{\text{Step 3}~~}\) List all possible factors of \(D\text{,}\) and compute the sum of each pair of factors. The factors must have opposite signs. (Complete each pair of factors below.)
Factors of \(D=30x^2\hphantom{0000}\) \(\hphantom{0000}\)Sum of Factors
\(-x\) \(\fillinmath{XXXXXXXXXX}\) \(\hphantom{0000}\)
\(-2x\) \(\fillinmath{XXXXXXXXXX}\) \(\hphantom{0000}\)
\(-3x\) \(\fillinmath{XXXXXXXXXX}\) \(\hphantom{0000}\)
The correct factors are:
\(\blert{\text{Step 4}~~}\)Enter the correct factors into the rectangle.
\(\blert{\text{Step 5}~~}\)Factor the top row of the rectangle, and write the result at the top. Finally, factor the bottom row, and write the result on the left. The correct factorization is
\begin{equation*} 4x^2+4x-3 = \hphantom{0000} \end{equation*}
boxes

Subsubsection Wrap-Up

Objectives.
In this Lesson we practiced the following skills:
  • Factoring a quadratic trinomial \(ax^2+bx+c\) by guess-and-check
  • Factoring a quadratic trinomial in two variables
  • Factoring out a binomial common factor
  • Combining techniques to factor completely
  • (Optional) Factoring a quadratic trinomial \(ax^2+bx+c\) by the box method
Questions.
  1. When is the guess-and-check method easier to use than the box method?
  2. In Activity 3, Problem 1, what is the common factor?
  3. In Activity 3, Problem 3a, what is the common factor?
  4. (Optional) When we use the box method, where do the factors of the trinomial appear?

Subsection Homework Preview

Exercises Exercises

1.
Factor \(3z^2-2z-5\) by the guess-and-check method.
2.
Factor completely.
  1. \(\displaystyle 2a^3+10a^2-72a\)
  2. \(\displaystyle 6x^4+27x^3y-54x^2y^2\)
Exercise Group.
\(\blert{\text{(Optional)}~~}\) Factor using the box method.
3.
\(2x^2-x-36\)
4.
\(4t^2-21t-18\)

Subsubsection Answers to Homework Preview

  1. \(\displaystyle (3z-5)(z+1)\)
    1. \(\displaystyle 2a(a+9)(a-4)\)
    2. \(\displaystyle 3x^2(2x-3y)(x+6y)\)
  3. \(\displaystyle (2x-9)(x+4)\)
  4. \(\displaystyle (4t+3)(t-6)\)

Exercises Homework 7.3

1.

Find each quotient by using the second law of exponents.
  1. \(\displaystyle \dfrac{a^6}{a^3}\)
  2. \(\displaystyle \dfrac{3^9}{3^4}\)
  3. \(\displaystyle \dfrac{z^6}{z^9}\)
Answer.
  1. \(\displaystyle a^3\)
  2. \(\displaystyle 3^6\)
  3. \(\displaystyle \dfrac{1}{z^3}\)

2.

Choose a value for the variable and evaluate to show that the following pairs of expressions are not equivalent.
  1. \(\displaystyle t^2 \cdot t^3,~~t^6\)
  2. \(\displaystyle \dfrac{v^8}{v^2},~~v^4\)
  3. \(\displaystyle \dfrac{n^3}{n^5},~~n^2\)

Exercise Group.

For Problems 3–5, divide.
3.
\(\dfrac{2x^3y}{8x^4y^5}\)
Answer.
\(\dfrac{1}{4xy^4} \)
4.
\(\dfrac{-12bx^4}{8bx^2}\)
5.
\(\dfrac{-15x^3y^2}{-3x^3y^4}\)
Answer.
\(\dfrac{5}{y^2} \)

Exercise Group.

For Problems 6–7, factor out a negative monomial.
6.
\(-b^2-bc-ab\)
7.
\(-4k^4+4k^2-2k\)
Answer.
\(-2k(2k^3-2k+1)\)

Exercise Group.

For Problems 8–11, factor out the greatest common factor.
8.
\(2x^4-4x^2+8\)
9.
\(16a^3b^3-12a^2b+8ab^2\)
Answer.
\(4ab(4a^2 b^2 - 3a +2b)\)
10.
\(9x^2-12x^5+3x^3\)
11.
\(14x^3y-35x^2y^2+21xy^3\)
Answer.
\(7xy(2x^2-5xy+3y^2)\)

Exercise Group.

For Problems 12–13, factor out the common factor.
12.
\(2x(x+6)-3(2x+6)\)
13.
\(3x^2(2x+3)-(2x+3)\)
Answer.
\((2x+3)(3x^2-1)\)

Exercise Group.

For Problems 14–16, factor by the guess-and-check method.
14.
\(2x^2+11x+5\)
15.
\(5t^2+7t+2\)
Answer.
\((5t+2)(t+1)\)
16.
\(3x^2-8x+5\)

Exercise Group.

For Problems 17–28, factor completely.
17.
\(2x^2+10x+12\)
Answer.
\(2(x+2)(x+3)\)
18.
\(4a^2b+12ab-7b\)
19.
\(4z^3+10z^2+6z\)
Answer.
\(2z(2z+3)(z+1)\)
20.
\(18a^2b-9ab-27b\)
21.
\(x^2-5xy+6y^2\)
Answer.
\((x-2y)(x-3y)\)
22.
\(x^2+4xy+4y^2\)
23.
\(x^2+4ax-77a^2\)
Answer.
\((x+11a)(x-7a)\)
24.
\(4x^3+12x^2y+8xy^2\)
25.
\(9a^3b+9a^2b^2-18ab^3\)
Answer.
\(9ab(a+2b)(a-b)\)
26.
\(2t^2-5st-3s^2\)
27.
\(4b^2y^2+5by+1\)
Answer.
\((4by+1)(by+1)\)
28.
\(12ab^2+15a^2b+3a^3\)

Exercise Group.

For Problems 29–34, solve the equation by factoring.
29.
\(3n^2-n=4\)
Answer.
\(-1\text{,}\) \(\dfrac{4}{3} \)
30.
\(11t=6t^2+3\)
31.
\(1=4y-4y^2\)
Answer.
\(\dfrac{1}{2}\text{,}\) \(\dfrac{1}{2} \)
32.
\(12z^2+26z=10\)
33.
\(y(3y+4)=4\)
Answer.
\(-2\text{,}\) \(\dfrac{2}{3} \)
34.
\((2x-1)(2x-1)=-1\)

35.

The cost \(C\) of producing a wool rug depends on the number of hours \(t\) it takes to weave it, where
\begin{equation*} C=3t^2-4t+100 \end{equation*}
How many hours did it take to weave a rug that costs $120?
Answer.
\(3\frac{1}{3} \) hrs

36.

Steve’s boat locker is 2 feet longer than twice its width. Find the dimensions of the locker if the 13-foot mast of Steve’s boat will just fit diagonally across the floor of the locker.
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